Question 936078
perimeter of a square = 4 * the side
area of a square = the side squared.


you get:


p = 4s
a = s^2


if the perimeter is 3 more than the area, then you get:


p = a + 3 which becomes p = s^2 + 3


you have p = s^2 + 3 and you have p = 4s


substitute s^2 + 3 for p in the second equation and you get:


s^2 + 3 = 4s


subtract 4s from both sides of this equation to get:


s^2 - 4s + 3 = 0


factor this equation to get:


(s-3) * (s+1) = 0


solve for s to get s = 3 or s = -1


s can't be negative, so s = 3


s^2 = 9


4s = 12


12 is 3 more than 9.


your solution is that the length of a side is equal to 3 and the area is equal to 9 and the perimeter is equal to 12.